JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field \(B =0.8\,T\). When released the radius of the loop starts shrinking at a constant rate of \(2\,cm ^{-1}\). The induced emf in the loop at an instant when the radius of the loop is \(10\,cm\) will be \(........mV\).
- A \(10\)
- B \(11\)
- C \(12\)
- D \(13\)
Answer & Solution
Correct Answer
(A) \(10\)
Step-by-step Solution
Detailed explanation
\(EMF =\frac{ d }{ dt }\left( B \pi r ^2\right)\) \(=2 B \pi r \frac{ dr }{ dt }=2 \times \pi \times 0.1 \times 0.8 \times 2 \times 10^{-2}\) \(=2 \pi \times 1.6= 1 0 . 0 6 \text { [round off } 1 0 . 0 6 = 1 0 ]\)
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